- 31 Jan '08 12:27

I experimented with it a bit and it seems likely that you got yourself a theorem. With all numbers I tried for n the result was always the form of a number ending in 0 and numbers that end in 0 are divisible by 5. But I can't get way to write your expresion in a product that has 10 in it.*Originally posted by FabianFnas***I found out that 1^5-1 is dividable with 5. So is 2^5-2 and 3^5-3.**

Is 2008^5-2008 also dividable with 5?

Is perhaps all n^5-n dividable with 5? Show me this is the case, or not.

This is all I got so far

n^5-n=n*(n^4-1)=n*(n^2+1)*(n^2-1)=n*(n^2+1)*(n+1)*(n-1)

Now n*(n+1) is the result of an arithmetic summation and I can't see if that help us or not but nonetheless it is interesting. And (n^2+1)*(n-1) maybe is something to though I don't know what...

I'll try for a bit longer and tell you about updates. - 31 Jan '08 13:12

Is 2008^5-2008 also dividable with 5? YES*Originally posted by FabianFnas***I found out that 1^5-1 is dividable with 5. So is 2^5-2 and 3^5-3.**

Is 2008^5-2008 also dividable with 5?

Is perhaps all n^5-n dividable with 5? Show me this is the case, or not.

2008^5 ends with the digit 8.

(since considering last digits 8*8 = 64, 4*8 = 32, 2*8 = 16, 6*8=48)

Subtracting 2008 gives us a number ending in zero and hence divisible by 5.

There are only 10 cases to consider; numbers ending in 0,1,2, ..9. Prove for each of these and your theorem is proven. - 31 Jan '08 15:46 / 1 editProof by induction that for all n >= 0, 5|(n^5 - n)

Base case

let n = 0

0^5 - 0 = 0 is divisible by 5 so result holds

Assume by inductive hypothesis result holds for n = k, ie; 5|(k^5-k)

Let n = k+1

(K+1)^5 - (k+1)

=k^5 +5k^4 +10k^3 + 10k^2 + 5k +1 - (k+1)

=k^5 +5(k^4 +2k^3 +2k^2 + k) - k

=(k^5 - k) + 5(k^4 +2k^3 +2k^2 + k) is divisible by 5 (inductive hypothesis)

Hence result holds for n = k+1

Therefore, by induction on n it is true that for all n >= 0, 5|(n^5 - n)

a similar argument can be used for n < 0 - 31 Jan '08 15:47 / 1 edit

If a not divisable by 5, hmm, what about when a is divisable by 5? Like 12345^5-12345 ?*Originally posted by Tricky Dicky***It's clearly true if the number you start with is divisible by 5.**

Fermat says a^4 = 1 (mod 5) if a not divisible by 5.

So a^5 = a (mod 5), whence the result.

But even if you don't know Fermat and what he says you can come to the correct result quite easy.

Edit: ... and now I see that Agerg has came up with a correct result. Well done! - 31 Jan '08 17:36 / 1 edit

well, clearly when a is divisible by 5, a=0 (mod 5) and so a^5=0 (mod 5)... so a^5-a = 0 (mod 5). The nonzero modulo classes are interesting because we can invoke Fermat's Little Theorem, a^p-1 = 1 (mod p) for any prime p. I think this is what Tricky Dicky was saying.*Originally posted by FabianFnas***If a not divisable by 5, hmm, what about when a is divisable by 5? Like 12345^5-12345 ?**

But even if you don't know Fermat and what he says you can come to the correct result quite easy.

Edit: ... and now I see that Agerg has came up with a correct result. Well done!

In fact, this proves that it is true for ALL prime number exponents, not just 5. Since a^(p-1)=1 (mod p), we see that a^p = a (mod p) and a^p - a = 0 (mod p).

So 12345^71-12345 is divisible by 71... and so forth. Fermat's "Little Theorem" is quite powerful

Edit: note, similarly to what Tricky Dicky said, a^p-1 = 1 (mod p) is only true when p does not divide a. but if it does, a^p = 0 (mod p) and a = 0 (mod p)... so the result a^p - a = 0 (mod p) still holds - 31 Jan '08 18:00I'm so not a mathematician but I think that this problem is simple and obvious. the number raised to any power with the number subtracted is obviously still a multiple of the number. no fancy formulas or anything else is needed, nor is it a theorem, at least any more than close the window there's a draft in here is a theorem.
- 31 Jan '08 18:38

n^p-n is still a multiple of n like you say but it is not obvious at all, at least not to me that n^p-n is a multiple of p.*Originally posted by coquette***I'm so not a mathematician but I think that this problem is simple and obvious. the number raised to any power with the number subtracted is obviously still a multiple of the number. no fancy formulas or anything else is needed, nor is it a theorem, at least any more than close the window there's a draft in here is a theorem.**

And sometimes on math the most difficult things to prove are the obvious ones. If you know calculus try to prove the intermediat value for yourself. Pretty self-evident and prety obvious. At first sight it even looks like not being a theorem at all. But just try to prove it for yourself. - 31 Jan '08 19:15

you're correct that this is obvious... what ISN'T obvious is that it's a multiple of the POWER you raised it to. that's the interesting fact: 3^71 - 3 is a multiple of 71 (in addition to the trivial fact it is a multiple of 3)*Originally posted by coquette***I'm so not a mathematician but I think that this problem is simple and obvious. the number raised to any power with the number subtracted is obviously still a multiple of the number. no fancy formulas or anything else is needed, nor is it a theorem, at least any more than close the window there's a draft in here is a theorem.** - 31 Jan '08 21:14

What does that ^ symbol mean?*Originally posted by Aetherael***you're correct that this is obvious... what ISN'T obvious is that it's a multiple of the POWER you raised it to. that's the interesting fact: 3^71 - 3 is a multiple of 71 (in addition to the trivial fact it is a multiple of 3)** - 01 Feb '08 00:19 / 3 edits

but we are looking for divisibility by the index...not the number raised to the index*Originally posted by coquette*

for example, consider 2^4-2...this is not divisible by 4 yet 2^5 -2 *is* divisible by 5...Yes m^n-m is certainly divisible by m but that is of no interest here....is it divisible by n??!

That is what we are interested in!